User:Egm4313.s12.team8.colocar/R3.7

Problem
Expand the series on both sides of (1)-(2) p.7-12b to verify equality.

$$ \displaystyle \sum_{j=2}^5 c_j j (j-1) x^{j-2} = \sum_{k=0}^{3} c_{k+2} (k+2) (k+1) x^{k} = \sum_{j=0}^{3} c_{j+2} (j+2) (j+1) {x^j}$$

$$ \displaystyle \sum_{j=1}^5 c_j \cdot j \cdot x^{j-1} = \sum_{k=0}^{4} c_{k+1} (k+1) x^{k} = \sum_{j=0}^{4} c_{j+1} (j+1) x^{j} $$

Solution
Negate middle portion of (1) and (2) as they display the transitional meaning to the function.

1  $$ \displaystyle \sum_{j=2}^5 c_j j (j-1) x^{j-2} = \sum_{j=0}^{3} c_{j+2} (j+2) (j+1) {x^j}$$

 $$ \displaystyle (c_2 2 (2-1) x^{2-2})+(c_3 3 (3-1) x^{3-2})+(c_4 4 (4-1) x^{4-2})+(c_5 5 (5-1) x^{5-2}) = (c_{0+2}(0+2)(0+1){x^0})+(c_{1+2}(1+2)(1+1){x^1})+(c_{2+2}(2+2)(2+1){x^2})+(c_{3+2}(3+2)(3+1){x^3}) $$

 $$ \displaystyle (c_2 2 (1) x^{0})+(c_3 3 (2) x^{1})+(c_4 4 (3) x^{2})+(c_5 5 (4) x^{3}) = (c_{2}(2)(1){x^0})+(c_{3}(3)(2){x^1})+(c_{4}(4)(3){x^2})+(c_{5}(5)(4){x^3}) $$

 $$ \displaystyle (2 c_2)+(6 c_3 x)+(12 c_4 x^{2})+(20 c_5 x^{3}) = 2c_2+ 6c_3 x + 12c_4 x^{2}+ 20c_5 x^{3} $$

2  $$ \displaystyle \sum_{j=1}^5 c_j \cdot j \cdot x^{j-1} = \sum_{j=0}^{4} c_{j+1} (j+1) x^{j} $$

 $$ \displaystyle (c_1 (1) x^{1-1})+(c_2 (2) x^{2-1})+(c_3 (3) x^{3-1})+(c_4 (4) x^{4-1})+(c_5 (5) x^{5-1}) = (c_{0+1} (0+1) x^{0})+(c_{1+1} (1+1) x^{1})+(c_{2+1} (2+1) x^{2})+(c_{3+1} (3+1) x^{3})+(c_{4+1} (4+1) x^{4}) $$

 $$ \displaystyle (c_1 x^{0})+(c_2 (2) x^{1})+(c_3 (3) x^{2})+(c_4 (4) x^{3})+(c_5 (5) x^{4}) = (c_{1} (1))+(c_{2} (2) x)+(c_{3} (3) x^{2})+(c_{4} (4) x^{3})+(c_{5} (5) x^{4}) $$

 $$ \displaystyle c_1+ 2c_2 x + 3c_3 x^{2} + 4c_4 x^{3}+ 5c_5 x^{4} = c_1+ 2c_2 x + 3c_3 x^{2} + 4c_4 x^{3}+ 5c_5 x^{4}  $$